Let $R$ be an integral domain. If $I \cap J = IJ$ for all ideals $I,J$ of $R,$ how do I show $R$ is a field? Hints will suffice. Thank you.
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Hint: Let $a\in R\setminus\{0\}$ and consider $I = J = Ra$.
Claudius
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Haha I got it from that thanks – Richard Martin Nov 19 '18 at 13:41
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1Let $r \in R.$ Then $r'a^{2} = ra, $ for some $r' \in R.$ Since $R$ is integral domain, $r = r'a \in Ra.$ So $1 \in Ra$ and hence $a$ is unit. – Alexy Vincenzo Nov 19 '18 at 13:56
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Don't you want to take $J$ as $R$ so the intersection will be $(a)$ and the product will be $R$? – John Douma Nov 19 '18 at 13:57
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@JohnDouma If $J=R$, then the product will be $Ra$, not $R$. – Claudius Nov 19 '18 at 13:58
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1@AlexyVincenzo Yes, this is, what I had in mind. Notice that you can assume $r=1$ to begin with. – Claudius Nov 19 '18 at 14:05
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@Claudius Thank you. I should have known it wasn't that easy because we wouldn't have needed the fact that $R$ is an integral domain. – John Douma Nov 19 '18 at 14:05
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Arithmetically: since ideal intersection $= \rm lcm$ for principal ideals, the hypothesis implies for all $\,a,b\neq 0\,$ we have $\,{\rm lcm}(a,b) = ab,\,$ so $\,\gcd(a,b)=1,\,$ i.e. $\,a,b$ have only unit common factors. Hence for $\,b=a\,$ the common factor $\,a\,$ of $\,a,b$ is a unit, so $\,a\neq 0\,\Rightarrow\, a\,$ is a unit, so $R$ is a field.
user26857
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Bill Dubuque
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Essentially we used: $,0\neq a$ is a unit $!\iff! \gcd(a,x) = 1!\iff! {\rm lcm}(a,x) = ax,\ $ for all $,x\neq 0.\ $ Recall that in general domains $\gcd$ and $\rm lcm$ are defined only up to unit factors (associates). – Bill Dubuque Nov 19 '18 at 16:43